Boundary of $L^1$ space Is there any rigorous or heuristic notion of boundary of $L^1$ that is studied? I mean something loosely like the collection of functions or distributions defined by
$$\left\{f\notin L^1: f_n\to f\quad\text{a.e.}\quad \text{as} \quad n\to \infty \quad \text{where} \quad f_n\in L^1\right\}$$
And what kind of characterizations or properties of this "surface" are known?
Edit: Changed to pointwise convergence.
 A: $L^1$ is Banach space, your set is therefore empty. You can also prove it in this way: for every $m,n$ we have
$$
\int_\mathbb{R}|f_n-f_m| \le \int_\mathbb{R}|f_n-f|+\int_\mathbb{R}|f-f_m|,
$$
i.e. $(f_n)$ is Cauchy sequence, and since $L^1$ is Banach space, there is some $g \in L^1$ such that $f_n \to g$. By uniqueness of the limit we conclude that $f=g \in L^1$.
A: I believe the set is just the set of all (non-integrable) measurable functions with $\sigma$-finite support.
In particular, if the space in question is $\sigma$-finite (like $\mathbf R$ with Lebesgue measure), then all measurable functions are pointwise limits of integrable functions.
Edit: cleaned up a bit.
In one direction:


*

*Clearly it is enough to show that for nonnegative functions;

*Let $f$ be a nonnegative measurable function, and $S_n$ an increasing sequence of sets of finite measure such that $\mathrm{supp}f\subseteq \bigcup_nS_n$.

*Let $A=\lbrace x\mid f(x)=\infty\rbrace$, $A_n=\lbrace x\mid f(x)<n\rbrace$.

*Then for each $n$ put $f_n(x)=f(x)$ on $A_n\cap S_n$, $f_n(x)=n$ on $A\cap S_n$, $f_n(x)=0$ otherwise.

*Then $f_n$ are integrable and $f_n\to f$ pointwise.


In the other direction, we show that the support of a pointwise limit of a sequence of integrable function has $\sigma$-finite support:


*

*Take an arbitrary sequence of integrable functions $f_n$

*Any integrable function $f_n$ has a $\sigma$-finite support $B_n=\bigcup_m\lbrace x\mid \lvert f_n(x)\rvert>1/m\rbrace$.

*The support of the limit of $f_n$ is contained in $\bigcup_n B_n$ (because if at some point none of the functions is nonzero, neither is the limit), and hence $\sigma$-finite, so we're done.

